If the coordinate direction angles for F3 = 700lb are alpha = 125 degrees, beta

Question

If the coordinate direction angles for F3 = 700lb are alpha = 125 degrees, beta = 40 degrees and gamma = 65 degrees, determine the magnitude of the resultant force acting on the eyebolt. (Figure 1). Determine coordinate direction angle alpha of the resultant force. Determine coordinate direction angle beta of the resultant force. Determine coordinate direction angle gamma of the resultant force.

If the coordinate direction angles for F3 = 700lb are alpha = 125 degrees, beta = 40 degrees and gamma = 65 degrees, determine the magnitude of the resultant force acting on the eyebolt. (Figure 1). Determine coordinate direction angle alpha of the resultant force. Determine coordinate direction angle beta of the resultant force. Determine coordinate direction angle gamma of the resultant force.

Explanation / Answer

force F1 as a vector: F1= 700cos30 i + 700sin30 j = 350sqrt(3) i + 350 j

F2= 600(3/5) j + 500(2/5) k = 360 j + 200 k

F3= 700(cos(125) i + cos(40) j + cos(65) k ) = -401.5 i + 536.2j + 295.83 k

resultant force (vector) => 204.72 i + 1246.2 j + 495.83 k

magnitude of this vectorial force => 1356.75 lb

direction cosines are:

alpha=> 81.32 degrees

beta=> 23.29 degrees

gamma=> 68.56 degrees

Hope it helps!

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